To the set 0 1 s and deduce that cantors inequality

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Unformatted text preview: t a successor. We know that whenever x a we have x a and since the
equation x a does not occur we must have x a whenever x a.
Now suppose that x a whenever x a. This condition tells us that there can’t exist a member x
of S for which x a and so a fails to be a successor.
9. If S is a well ordered set and a S, then we saw that a is a limit member of S if a is not the least member of
S and a has no predecessor. Prove that if a well ordered set S has a limit member then S has a nonempty
subset E that has no limit member and no largest member.
Suppose that a well ordered set S has a limit member. We define q to be the least limit member of S
and we define E x S x q . We see at once that E is nonempty and has no limit member.
Now given any member x of E, it follows from Exercise 8 and the fact that x q that x E and so
E has no largest member.
10. Give an example of a totally ordered set S that is not well ordered even though every member of S has a
successor.
The system Z of integers has the desired properties. Exercises on Zorn’s Lemma
1. a. A set S of real numbers is said to be linearly independent over the set Q of rational numbers if for every
positive integer n and every choice of members x 1 , , x n of the set S and rational numbers r 1 , , r n , the
condition
r1x1 rnxn 0
cannot hold unless all of the numbers r 1 , , r n are zero. Prove that if is the family of subsets of R that
are linearly independent over Q and , is partially ordered by the relation , then has a maximal
member.
We need to show that every chain in the partially ordered set has an upper bound. Suppose
that is a chain in . Each member of is linearly independent over the set Q of rational
numbers and if A and B are any two members of then either A B or B A. We define
U Þ and we observe that U is a set of numbers and A U whenever A
. All we need to
do is show that U is linearly independent.
Suppose that n is a positive integer and that x 1 , x 2 , , x n belong to . We see easily that some
member A of contains all of the numbers x 1 , x 2 , , x n . Since A is linearly independent, no
linear combination
r1x1 rnxn
with rational coefficients can be zero unless all of the coefficients are zero. This completes the
proof that U
and so the chain has an upper bound in . 69 b. Prove that if H is a maximal member of the family that was defined in part a then every real number
can be expressed uniquely in the form r 1 x 1 r n x n .
2. An additive subgroup of the system R of real numbers is defined to be a nonempty subset G of R with the
property that whenever x and y belong to G, then so do the numbers x y and x y.
a. Which of the following subsets of R are additive subgroups of R?
These simple exercises have appeared before.
i. The set 0 .
ii. The set 1 .
iii. The set 1, 0, 1 . iv. The set Z of all integers.
v. The set Q of all rational numbers.
vi. The set R Q of all irrational numbers. vii. The empty set .
b. Given that G is an...
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